# Reimann integral question

• July 8th 2011, 10:35 AM
CountingPenguins
Reimann integral question
Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
• July 8th 2011, 10:53 AM
girdav
Re: Reimann integral question
What we suppose works for all Riemann-integrable function $g$, and we only know that $\int_a^b fg =0$.
• July 8th 2011, 11:09 AM
Plato
Re: Reimann integral question
Quote:

Originally Posted by CountingPenguins
Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

From the given do you know that $\int_a^b {f^2 } = 0~?$

Is that possible if $f(t)\not=0$ for some $t\in[a,b]~?$
• July 8th 2011, 12:29 PM
CountingPenguins
Re: Reimann integral question
f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.
• July 8th 2011, 12:45 PM
Plato
Re: Reimann integral question
Did you see reply #3. That answers the question.
• July 10th 2011, 01:49 PM
Also sprach Zarathustra
Re: Reimann integral question
Quote:

Originally Posted by CountingPenguins
Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both?

I came up with the following solution, I hope it is not so wrong. (Itwasntme)

Suppose that $f$ is not $0$ for all $x\in[a,b]$, then there exists $x_0\in[a,b]$ such that $f(x_0)>0$ (If the is no such $x_0$, there must be $x_1\in[a,b]$ such that $f(x_1)<0$.In that case we will look at $-f$).

f is continuous function therefor exists $\delta>0$ for which $f(x)>\epsilon>0$ for all $x\in (x_0-\delta,x_0+\delta)$.

Now we define $h(x)$-'staircase'-characteristic function of open interval $(x_0-\delta,x_0+\delta)$.

$h(x)=\left\{\begin{matrix}1 & x\in (x_0-\delta,x_0+\delta) \\ 0 & \text{otherwise} \end{matrix}\righ$

Hence, we will get:

$0=\int_{a}^{b}f(x)h(x)dx=\int_{a}^{x_0-\delta}f(x)h(x)dx+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+\int_{x_0+\delta}^{ b}f(x)h(x)dx=0+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+0>2\epsilon\delta$.

So, we got a contradiction to the fact that if $f$ is continuous on $[a,b]$ and h is staircase function $\int_{a}^{b}f(x)h(x)dx=0 \Rightarrow f\equiv 0$.

But we also know that if $g$ is integrable function on $[a,b]$, so for every $\epsilon_0>0$ exists staircase function $h$ so that

$\int_{a}^{b}|g(x)-h(x)|<\epsilon_0$

Combining these two results we will conclude the needed.
• July 10th 2011, 03:35 PM
Plato
Re: Reimann integral question
Quote:

Originally Posted by Also sprach Zarathustra
I came up with the following solution, I hope it is not so wrong.

Your solution is correct.
Note that $f \text{ is continuous on }[a,b]$ so as such that means that both $f~\&~f^2$ is are integrable on $[a,b]$.

By the given $\int_a^b {f \cdot f} = \int_a^b {f^2 } = 0$.
But by your reasoning that must imply that $f \equiv 0$.